Determine the output impedance of the circuit

[brainbaby]

TL;DR Summary

To find output impedance of ltspice circuit with three ac sine source.

Hi friends,

I have taken three sine AC sources and have joined them in series. I want to find the output impedance of this circuit.

Formula that I am using is...Output impedance = (Open circuit voltage - Output voltage with load R1) / Output current with load (R1)
Load resistance = 1 k ohm.

The problem arises when I am trying to find the open circuit voltage, as there is no change in output voltage (RMS = 893.42 mV):-
i) When load resistance is 999999 and
ii) When load resistance is 1k.

Where I could be wrong?
Please find .asc file in the attachment.

 

[BvU]

hi baby,

Can you do the exercise with one single voltage source first ?

(the straight answer isn't useful for you at this moment -- is my estimate)

 

[berkeman]

Summary: To find output impedance of ltspice circuit with three ac sine source.

Where I could be wrong?

Please double-click on one of the voltage sources and show us the model. You may be able to answer your own question by just looking at the default model you are using...

 

[brainbaby]

Can you do the exercise with one single voltage source first ?

No luck!

 

[berkeman]

That's the whole voltage source model? What simulator software is that?

Let me check my MicroCAP SPICE package...

 

[brainbaby]

@berkeman

Its ltspice...This might help you!

 

[berkeman]

Yes, that's better. So if you leave that "Series Resistance" box blank, what do you think it defaults to?

In MicroCAP-9 you can see what it defaults to...

 

[brainbaby]

Yes, that's better. So if you leave that "Series Resistance" box blank, what do you think it defaults to?

I think so it refers to the internal resistance of the voltage source.

 

[berkeman]

Yes. Isn't that what you are trying to determine with your simulation? Or did I misread your original post?

 

[brainbaby]

Yes. Isn't that what you are trying to determine with your simulation? Or did I misread your original post?

A bit yes...
I will reiterate again..
My question is that I have taken two values of load resistance for this circuit ...
one is 1k ohms and other is 99999 ohms (open circuit voltage).

My aim is to find the output impedance...

The output impedance, is equal to the difference of the open circuit voltage and output voltage when load resistance is present(i.e 1k) over current flowing through 1k load resistance.

But after simulating I found that there is no significant change in the output voltage , in other words in this case the output voltage is independent of the load resistance value...which I don't think so is theoretically valid...

 

[berkeman]

The output impedance, is equal to the difference of the open circuit voltage and output voltage when load resistance is present(i.e 1k) over current flowing through 1k load resistance.

It is also equal to the sum of the internal resistances of your series-connected voltage sources, right? Where else would the output impedance come from in your circuit?

And if you take the value of Rs from my voltage source SPICE model, what would be the total output resistance for 3 of them connected together in series? Can you then say why the load resistance doesn't matter much in changing the output voltages?

 

[BvU]

No luck!

I'm not unhappy with this result at all: it shows that the output of an ideal (without 'parasitic' series resistance *) voltage source does not depend on the load. Ergo: no output impedance ...

But I see @berkeman is already helping you out. Kudos !

*) or capacitance

 

[essenmein]

But after simulating I found that there is no significant change in the output voltage , in other words in this case the output voltage is independent of the load resistance value...which I don't think so is theoretically valid...

The thing about simulation tools is they are idealized environments. In the real world a voltage source with zero output impedance (ie the output voltage is independent of the load resistance) does not exist, you can have very low output impedances, but not zero.

However when simply solving equations, which is all the simulation tool is doing, you can most certainly set the source impedance to zero, just like you can set a load resistance to zero, which will then make the computer complain (or crash if it's poorly implemented code).

 

[berkeman]

My aim is to find the output impedance...

BTW, the output impedance of real signal generators (lab instruments like the HP 33120 Signal Generator / AFG) is generally 50 Ohms on purpose. That makes the interfacing with other 50 Ohm instruments more straightforward.

Quiz Question for the OP -- So if I monitor the output of the 33120 with my oscilloscope in the lab, and set my 33120 AC Voltage Output amplitude to 1Vpp, what voltage do I see when I use a high impedance 'scope probe to check that output voltage? What voltage do I measure on my 'scope when I use a direct coax to my 'scope when I set the input coupling to 50 Ohms versus DC 1MegOhm?



https://www.us-instrument.com/conte...nerator-arbitrary-waveform-generator_550.jpeg

 

[brainbaby]

It is also equal to the sum of the internal resistances of your series-connected voltage sources, right? Where else would the output impedance come from in your circu

Yes, in this case the output resistance will be = voltage produced by voltage source (open circuit condition) / Short circuit current( output terminal shorted).
but this method is not feasible with spice software, also practically it is not an advisable method to measure the output resistance as it can upset the operating conditions of the circuit.

The method that seemed to be most suitable for me was that which I have mentioned a couple of times in my previous posts (post 1 and 10).

 

[brainbaby]

Quiz Question for the OP -- So if I monitor the output of the 33120 with my oscilloscope in the lab, and set my 33120 AC Voltage Output amplitude to 1Vpp, what voltage do I see when I use a hig

Input impedance of the next stage is always kept high as it won't load the previous circuit. The output voltage measured with a high impedance probe will be near to 1 Vpp. In other words minimum current should flow in the measuring device so that it do not disturb the output voltage that is to be measured.

This was equivalent to the case where I kept load resistance as 99999 ohms.

 

[brainbaby]

Can you then say why the load resistance doesn't matter much in changing the output voltages?

Yes I agree that the output impedance may also vary according to the load impedance across the output terminals. But I don't see any change in the output parameters for two different values of load resistances.

Simulation data
When R1=99999 and 1K (No Change in any of the parameters)

Vo

V4
20.569nV Average voltage
424.13mV RMS voltage

V3
19.775µV
702.99mV

V1
918.24nV
352.31mVI(R1)

V4
45.145pAa Average current
883.79nA RMS value

V3
45.145pA
883.79nA

V1
45.145pA
883.79nA

 

[berkeman]

Input impedance of the next stage is always kept high as it won't load the previous circuit. The output voltage measured with a high impedance probe will be near to 1 Vpp.

No. Instruments that have a 50 Ohm output impedance expect to drive a 50 Ohm load. So you get 1Vpp when you connect the signal generator to a 50 Ohm load.

If you drive the output into a high impedance, you see 2Vpp when the signal generator is set to 1Vpp. Can you say why that is?

 

[berkeman]

Yes I agree that the output impedance may also vary according to the load impedance across the output terminals.

No. that is not generally true for electronic sources. For batteries, it may be true that the output impedance depends on the current being drawn from it (and the state of the battery's charge, and the age of the battery, etc.).

But I don't see any change in the output parameters for two different values of load resistances.

Most likely because the output impedance of those default voltage sources in your simulator are close to mine, which is in the low miliOhm range. So essentially zero output impedance by default.

Instead, use a single voltage source in your simulation and set its output resistance to 50 Ohms. Then try a few load impedances to plot how the output voltage from the source to the load varies with load resistance. That should help you to understand what is going on. Have fun!

 

[brainbaby]

Couple of questions more..

Instruments that have a 50 Ohm output impedance expect to drive a 50 Ohm load. So you get 1Vpp when you connect the signal generator to a 50 Ohm load.

If you drive the output into a high impedance, you see 2Vpp when the signal generator is set to 1Vpp. Can you say why that is?

What I read was that for optimum voltage transfer between two stages the output impedance of the first circuit should be much lower than the input impedance of the second If this condition is not met and the source impedance is equal to the load impedance loss of signal occurs approx. 50%. But then if "50 Ohm output impedance expect to drive a 50 Ohm load." then the voltage transfer to load should be nearly half of the source voltage and when output is driven by a high impedance, we must seen 1Vpp as a result.

But maximum power theorem states for max power transfer Rs (source resistance = load resistance)

? Didn't got.

Most likely because the output impedance of those default voltage sources in your simulator are close to mine, which is in the low miliOhm range. So essentially zero output impedance by default

Got it.

nstead, use a single voltage source in your simulation and set its output resistance to 50 Ohms. Then try a few load impedances to plot how the output voltage from the source to the load varies with load resistance. That should help you to understand what is going on.

I have selected the series resistance as 50 ohm and simulated different values of load resistance with only one source what I observed is :
When very minimum or zero output impedance is selected for the circuit the output voltage becomes independent of the load resistance. may not be totally independent but the changes in output voltage are very negligible. Why is it so?

It is also equal to the sum of the internal resistances of your series-connected voltage sources, right?

Can I take the output impedance of circuit containing 3 sources as 150 ohms?

 

[berkeman]

What I read was that for optimum voltage transfer between two stages the output impedance of the first circuit should be much lower than the input impedance of the second If this condition is not met and the source impedance is equal to the load impedance loss of signal occurs approx. 50%. But then if "50 Ohm output impedance expect to drive a 50 Ohm load." then the voltage transfer to load should be nearly half of the source voltage and when output is driven by a high impedance, we must seen 1Vpp as a result.

But maximum power theorem states for max power transfer Rs (source resistance = load resistance)

? Didn't got.

When dealing with controlled impedance circuits, you will match the source and load impedances. That gives you the best power transfer, and is the easiest to deal with.

You might want a high input impedance looking into an amplifier when you are trying to optimize your voltage signal-to-noise ratio, like in sensor circuits. As soon as you can in the amplifier chain, though, you will start matching output and input impedances.

Can I take the output impedance of circuit containing 3 sources as 150 ohms?

Yes, the internal resistances of the sources add in series when the voltage sources are connected in series.

 

[Baluncore]

LTspice models voltage sources as having Rser = zero, unless you specify otherwise.
The source resistance of the three perfect voltage sources in series is zero.

The amplitude of the output voltage will halve when the load resistance is equal to the source resistance. When I change the load from 1MEG ohm to 1u0 ohm I see no change in the output amplitude, which confirms the three perfect sources in series have zero source impedance.

 

[brainbaby]

Another thing...
I have expected my output to be a modulated output.
V1 is the baseband signal. V3 is the carrier and V4 is a random signal.

As you can see the output wave form doesn't look like a modulated signal.

How to correct this??

LTspice help.

 

[berkeman]

Another thing...
I have expected my output to be a modulated output.
V1 is the baseband signal. V3 is the carrier and V4 is a random signal.

As you can see the output wave form doesn't look like a modulated signal.

How to correct this??

LTspice help.

Adding signals does not create modulation. You need to multiply them.

 

[Baluncore]

With LTspice you can examine the frequency spectrum of the output signal.
Select the time plot of the signal you want to analyse, then the view menu, select FFT.
I see your three sinewave signals without any modulation.

 

[Baluncore]

Attached here is an LTspice AM modulator, with added white noise.